Advertisements
Advertisements
प्रश्न
Evaluate:
Advertisements
उत्तर
\[\int\left( \frac{\cos 2x + 2 \sin^2 x}{\sin^2 x} \right)dx\]
\[ = \int\left( \frac{1 - 2 \sin^2 x + 2 \sin^2 x}{\sin^2 x} \right)dx \left[ \because \cos 2x = 1 - 2 \sin^2 x \right]\]
\[ = \int {cosec}^2\text{ x dx}\]
\[ = - \cot x + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate: `int (1+logx)/(x(2+logx)(3+logx))dx`
Evaluate: `int1/(xlogxlog(logx))dx`
Evaluate : `int1/(3+5cosx)dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate: `intsinsqrtx/sqrtxdx`
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
`int 1/(1 + cos x)` dx = _____
A) `tan(x/2) + c`
B) `2 tan (x/2) + c`
C) -`cot (x/2) + c`
D) -2 `cot (x/2)` + c
Evaluate of the following integral:
(i) \[\int x^4 dx\]
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate :
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate the following integral:
Evaluate :
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" d"x"`.
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
Evaluate the following:
`int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`
Evaluate the following:
`int "dt"/sqrt(3"t" - 2"t"^2)`
Find: `int (dx)/sqrt(3 - 2x - x^2)`
The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is
